3.29 \(\int \sec ^2(e+f x) (2-3 \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=19 \[ -\frac {\tan (e+f x) \sec ^2(e+f x)}{f} \]

[Out]

-sec(f*x+e)^2*tan(f*x+e)/f

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {4043} \[ -\frac {\tan (e+f x) \sec ^2(e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2*(2 - 3*Sec[e + f*x]^2),x]

[Out]

-((Sec[e + f*x]^2*Tan[e + f*x])/f)

Rule 4043

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] /; FreeQ[{b, e, f, A, C, m}, x] && EqQ[C*m + A*(m + 1), 0]

Rubi steps

\begin {align*} \int \sec ^2(e+f x) \left (2-3 \sec ^2(e+f x)\right ) \, dx &=-\frac {\sec ^2(e+f x) \tan (e+f x)}{f}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 19, normalized size = 1.00 \[ -\frac {\tan (e+f x) \sec ^2(e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^2*(2 - 3*Sec[e + f*x]^2),x]

[Out]

-((Sec[e + f*x]^2*Tan[e + f*x])/f)

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fricas [A]  time = 0.44, size = 19, normalized size = 1.00 \[ -\frac {\sin \left (f x + e\right )}{f \cos \left (f x + e\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(2-3*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

-sin(f*x + e)/(f*cos(f*x + e)^3)

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giac [A]  time = 0.44, size = 22, normalized size = 1.16 \[ -\frac {\tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(2-3*sec(f*x+e)^2),x, algorithm="giac")

[Out]

-(tan(f*x + e)^3 + tan(f*x + e))/f

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maple [A]  time = 1.08, size = 34, normalized size = 1.79 \[ \frac {2 \tan \left (f x +e \right )+3 \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^2*(2-3*sec(f*x+e)^2),x)

[Out]

1/f*(2*tan(f*x+e)+3*(-2/3-1/3*sec(f*x+e)^2)*tan(f*x+e))

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maxima [A]  time = 0.33, size = 20, normalized size = 1.05 \[ -\frac {\tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(2-3*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

-(tan(f*x + e)^3 + tan(f*x + e))/f

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mupad [B]  time = 2.40, size = 21, normalized size = 1.11 \[ -\frac {\mathrm {tan}\left (e+f\,x\right )\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3/cos(e + f*x)^2 - 2)/cos(e + f*x)^2,x)

[Out]

-(tan(e + f*x)*(tan(e + f*x)^2 + 1))/f

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- 2 \sec ^{2}{\left (e + f x \right )}\right )\, dx - \int 3 \sec ^{4}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**2*(2-3*sec(f*x+e)**2),x)

[Out]

-Integral(-2*sec(e + f*x)**2, x) - Integral(3*sec(e + f*x)**4, x)

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